Question

if sin A= 3/4, calculate cos A and tan A​

Answer 1

sin A=3/4

BC/AC=3/4

Let

BC=3x

AC=4x

(AC)^2=(AB)^2+(BC)^2

(4x)^2=(AB)^2+(3x)^2

16x^2-9x^2=(AB)^2

AB=root(7x)

cos A= AC/AC

=root(7x)/4x

=root(7)/4

tan A= BC/AB

=3x/root(7x)

=3/root(7)

Answer 2

Given that,

• $\rm { sin \: A = \dfrac{3}{4} }$

We have to find,

• $\rm { cos \: A \: and \: tan \: A }$

Solution,

Here, we know that

$\rm { sin \: A =\dfrac{ 3}{4} = \dfrac{ Perpendicular}{Hypotenuse} }$

Hence,

• $\rm { Perpendicular = 3}$
• $\rm { Hypotenuse = 4}$

_____________

Applying pythagoras property-

$\rm\red { {H}^{2} = {B}^{2} + {P}^{2} }$

$\rm { \leadsto {B}^{2} = {H}^{2} - {P}^{2} }$

$\rm { \leadsto {B}^{2} = {4}^{2} - {3}^{2} }$

$\rm { \leadsto {B}^{2} = 16 - 9 = 7}$

$\rm\blue { \leadsto B = \sqrt{7} }$

________________

• $\rm { Hypotenuse = 4 }$
• $\rm { Base = \sqrt{7} }$
• $\rm { Perpendicular = 3 }$

$\leadsto \rm\red{ cos \: A = \dfrac{ Base}{ Hypotenuse} = \dfrac{ \sqrt{7} }{4} }$

$\:$

$\leadsto \rm\red{ tan \: A = \dfrac{ Perpendicular}{ Base} = \dfrac{ 3 }{\sqrt{7}} }$